Integration and the Binomial Theorem (a simple question) (1 Viewer)

[Equilibrium1]

This may be a little hard to explain but, if it helps, I am referring to Q56(b) on Pg 210 of the 3 Unit Fitzpatrick book or, alternatively, the bottom of page 202 in the 3 unit Cambridge Textbook (Year 12).

Anyway, when asked to integrate a binomial expression, such as (1 + x)^n and represent it in the form of: [2^(n+1) - 1] / [n +1]

I'm having trouble obtaining the -1 in the numerator. I checked the working out and its because I did not add C (the constant). This makes sense, however, when you integrate (1 + x)^n (the LHS) and the expanded/sigma form (the RHS) wouldn't the C's simply cancel out?

Because in the working out in Cambridge and the Fitzpatrick solutions, they only add C on the RHS (i.e. the expanded/sigma form). Why don't they add C on the LHS? (since that side is integrated too).

I know how to work it out, but I just don't understand the part about the constant =/

Thankyou very much, I really appreciate it.

 

[nightweaver066]

This may be a little hard to explain but, if it helps, I am referring to Q56(b) on Pg 210 of the 3 Unit Fitzpatrick book or, alternatively, the bottom of page 202 in the 3 unit Cambridge Textbook (Year 12).

Anyway, when asked to integrate a binomial expression, such as (1 + x)^n and represent it in the form of: [2^(n+1) - 1] / [n +1]

I'm having trouble obtaining the -1 in the numerator. I checked the working out and its because I did not add C (the constant). This makes sense, however, when you integrate (1 + x)^n (the LHS) and the expanded/sigma form (the RHS) wouldn't the C's simply cancel out?

Because in the working out in Cambridge and the Fitzpatrick solutions, they only add C on the RHS (i.e. the expanded/sigma form). Why don't they add C on the LHS? (since that side is integrated too).

I know how to work it out, but I just don't understand the part about the constant =/

Thankyou very much, I really appreciate it.

After you integrate, leave the c there. Sub x = 0 to obtain c and you get your answer

 

[Equilibrium1]

Hey thanks for your answer,

I know how to work it out, i.e. sub x=0 etc...

I just dont understand why they dont add C on the LHS when they've added it on the RHS.

 

[nightweaver066]

Hey thanks for your answer,

I know how to work it out, i.e. sub x=0 etc...

I just dont understand why they dont add C on the LHS when they've added it on the RHS.

You dont know what the constants are on each side, you can't assume theyre both c.

To prevent this problem, just add c to one side which can account for the constants on both side.

 

[Carrotsticks]

When you integrate both sides, you will get constants on both sides. Let them be A and B. We move either one to the other side, lets say we move the A to the B. So B becomes B-A. But this is another positive or negatice (doesn't matter). So we call this new constant C. Hence only one constant in total.

The constants of integration dont necessarily cancel each other out because they may not be the same.